Solid State

To find the value of $ X $ in the metal deficient oxide sample $ M_x Y_2 O_4 $, where $ M $ is present in both +2 and +3 oxidation states and $ Y $ is in the +3 oxidation state, and given that the fraction of $ M^{2+} $ ions present in $ M $ is $\frac{1}{3}$, follow these steps:

1. Charge Balance Equation:

The total charge contributed by the cations ($ M $ and $ Y $) must balance the total negative charge contributed by the oxide ions $ O $:

$ x(M^{2+} + M^{3+}) + 2(Y^{3+}) = 4(-2) $

1. Oxidation States and Charges:

• Let the total number of $ M $ ions be $ x $.


• Given $\frac{1}{3}$ of $ M $ ions are $ M^{2+} $, the remaining $\frac{2}{3}$ are $ M^{3+} $.

1. Number of Ions:

• Number of $ M^{2+} $ ions = $\frac{x}{3}$.


• Number of $ M^{3+} $ ions = $\frac{2x}{3}$.

1. Total Positive Charge:

• Charge from $ M^{2+} $ ions = $ \left(\frac{x}{3}\right) \times 2 = \frac{2x}{3} $.


• Charge from $ M^{3+} $ ions = $ \left(\frac{2x}{3}\right) \times 3 = 2x $.


• Charge from $ Y^{3+} $ ions = $ 2 \times 3 = 6 $.

1. Total Positive Charge Calculation:

$ \frac{2x}{3} + 2x + 6 $

1. Total Negative Charge:

• Charge from 4 oxygen ions $ (O^{2-}) $:

$ 4 \times (-2) = -8 $

1. Setting Up the Charge Balance:

$ \frac{2x}{3} + 2x + 6 = 8 $

1. Solving for $ x $:

$ \frac{2x}{3} + 2x + 6 = 8 $

$ \frac{2x}{3} + 2x = 2 $

$ \frac{2x + 6x}{3} = 2 $

$ \frac{8x}{3} = 2 $

$ 8x = 6 $

$ x = \frac{6}{8} $

$ x = 0.75 $

Thus, the value of $ X $ is $ \boxed{0.75} $.

Answer:

Option D: 0.75

1. Atomic radii relations :

Given,

$ r_{y} = \frac{8}{\sqrt{3}} r_{x} \tag{1} $ ......(1)

$ r_{z} = \frac{\sqrt{3}}{2} r_{y} \tag{2} $ ........(2)

Substituting Eq. (1) into Eq. (2), we get :

$ r_{z} = \frac{\sqrt{3}}{2} \times \frac{8}{\sqrt{3}} r_{x} = 4 r_{x} \tag{3} $ .......(3)

2. Edge lengths of the unit cells :

The edge lengths (L) of the unit cells for x, y, and z are given by :

$ L_x = 2 \sqrt{2} r_x \quad \text{(for FCC)} \tag{4} $ .......(4)

$ L_y = 4 r_y = \frac{32}{\sqrt{3}} r_x \quad \text{(for BCC, using Eq. (1))} \tag{5} $ .......(5)

$ L_z = 2 r_z = 8 r_x \quad \text{(for simple cubic, using Eq. (3))} \tag{6} $ .......(6)

3. Comparing the edge lengths :

From Eq. (4), (5) and (6), we find :

$ L_y > L_z > L_x \tag{7} $

This means Option B : $L_y > L_z$ is correct.

4. Packing Efficiencies :

The packing efficiencies for FCC, BCC, and simple cubic are calculated as follows :

- FCC :

$ PE_{fcc} = \frac{4 \times \frac{4}{3} \pi r_x^3}{(L_x)^3} = \frac{\pi}{\sqrt{2}} \tag{8} $

- BCC :

$ PE_{bcc} = \frac{2 \times \frac{4}{3} \pi r_y^3}{(L_y)^3} = \frac{3 \pi \sqrt{3}}{32} \tag{9} $

- Simple Cubic :

$ PE_{sc} = \frac{1 \times \frac{4}{3} \pi r_z^3}{(L_z)^3} = \frac{\pi}{6} \tag{10} $

Comparing these packing efficiencies, we find :

$ PE_{fcc} > PE_{bcc} > PE_{sc} \tag{11} $

This means Option A : Packing efficiency of unit cell of $x > y > z$ is correct.

5. Calculating Densities :

The densities of x and y are given by the formula $d = \frac{Z \cdot M}{N_a \cdot L^3}$, where $Z$ is the number of atoms in the unit cell, $M$ is the molar mass of the metal, $N_a$ is Avogadro's number, and $L$ is the edge length of the unit cell.

- For x (FCC), $Z=4$ :

$ d_x = \frac{4 M_x}{N_a (L_x)^3} \tag{12} $

- For y (BCC), $Z=2$ :

$ d_y = \frac{2 M_y}{N_a (L_y)^3} \tag{13} $

$M_z=\frac{3}{2} M_y$ and $M_z=3 M_x$. This implies that $M_x=\frac{M_y}{2}$. Thus, $2M_x=M_y$, we can write :

$ \frac{d_x}{d_y} = \frac{2 M_x}{M_y} \times \left(\frac{L_y}{L_x}\right)^3 \tag{14} $ .......(7)

Substituting the given relation $2 M_x = M_y$ into Eq. (7), we get :

$ \frac{d_x}{d_y} = 2 \times 1 \times \left(\frac{32/3}{2 \sqrt{2}}\right)^3 \tag{15} $ ..........(8)

After simplifying Eq. (8), we find $d_x > d_y$.

This means Option D : Density of $x > y$ is correct.

So, the correct answers are Options A, B, and D.

Atom $X$ occupies FCC lattice sites as well as alternate tetrahedral voids of FCC.

In FCC, tetrahedral voids are 8 (in a unit cell)

Hence

Atom $X$ in a unit cell (FCC lattice sites) $=8 \times \frac{1}{8}+6 \times \frac{1}{2}=4$

Atom $X$ in a unit cell (in T.V.) $=\frac{1}{2} \times 8=4$

Total atom $\mathrm{X}$ in one unit cell $=8$

For relation between a and r, since T.V. forms at $\frac{1}{4}$ th of body diagonal,

$ \begin{aligned} & \frac{a \sqrt{3}}{4}=2 r \\\\ &\Rightarrow a=\frac{8 r}{\sqrt{3}} \end{aligned} $

Packing efficiency

$=\frac{8 \times \frac{4}{3} \pi r^3}{a^3} \times 100=\frac{8 \times \frac{4}{3} \pi r^3}{\left(\frac{8 r}{\sqrt{3}}\right)^3} \times 100 \simeq 35 \%$

Radius of cation X = r₊

Number of cation X present = $6 \times {1 \over 2} = 3$

Radius of anion Y = r_$-$

Number of anion Y present = ${1 \over 8} \times 8 = 1$

Side of unit cell = a

As two anions are touching each other, so side of unit cell,

a = 2r_$-$

Volume of unit cell = a³ = (2r_$-$)³ = 8r$_ - ^3$

Volume of cations X = Number of cation $\times$ Volume of one cation = 3 $\times$ ${4 \over 3}\pi r_ + ^3$

Volume of anions Y = Number of anion $\times$ Volume of one anion = 1 $\times$ ${4 \over 3}\pi r_ - ^3$

Radius ratio for close packed structure, ${{{r_ + }} \over {{r_ - }}} = 0.414$

Packing efficiency = ${{Volume\,of\,cation \times Number\,of\,cations + Volume\,of\,anion \times Number\,of\,anions} \over {Volume\,of\,unit\,cell}} \times 100$

Packing fraction = ${{Packing\,efficiency} \over {100}}$

$ = {{1 \times {4 \over 3}\pi r_ - ^3 + 3 \times {4 \over 3}\pi r_ + ^3} \over {8r_ - ^3}}$

$ = {{4\pi } \over {3 \times 8}}\left[ {{{r_ - ^3} \over {r_ - ^3}} + {{3r_ + ^3} \over {r_ - ^3}}} \right] = {\pi \over 6}[1 + 3{(0.414)^3}]$

$ = {{3.14} \over 6}(1 + 3 \times 0.0710) = {{3.14} \over 6} \times 1.213 = 0.63$

(a) The empirical formula of the compound :

Contribution of M and X :

${M_{\left( {2 \times {1 \over 2}} \right)}}{X_{\left( {4 \times {1 \over 4}} \right)}} \Rightarrow MX$

(b) Coordination number of both M and X is 8.

(c) Distance between M and X

$ = \sqrt {{{{a^2}} \over 4} + {{{a^2}} \over 2}} = \sqrt {{3 \over 4}} a = {{\sqrt 3 } \over 2}a$

$ \Rightarrow $ 0.866 a

(d) ${r_M}\,:\,{r_X} = (\sqrt 3 - 1)\,:\,1 \Rightarrow 0.732:1$, thus statement (d) is incorrect.

Coordination number cannot be 12, for any atom in the topmost layer, as there is no layer above it. Thus, each atom is in contact with six atoms in the same layer and three atoms from the layer below it.

For cubic close packing, we have

Packing fraction = ${{Volume\,of\,four\,spheres\,in\,the\,unit\,cell} \over {Total\,volume\,of\,the\,unit\,cell}}$

$ = 4 \times {{(4/3)\pi {r^3}} \over {16\sqrt 2 {r^3}}} = {\pi \over {3\sqrt 2 }} = 0.74 = 74\% $

In fcc unit cell, the effective number of atoms is

One atom at each corner = 8 corner atoms $ \times {1 \over 8} = 1$

Atoms at each of the six face centres :

6 face centered atoms $ \times {1 \over 2} = 3$.

Number of octahedral voids = 4.

Number of tetrahedral voids = 8.

Therefore, per atom, there is one octahedral void and two tetrahedral voids.

In fcc (or ccp), the unit edge length is given by

$\sqrt 2 a = 4r \Rightarrow a = {{4r} \over {\sqrt 2 }} = 2\sqrt 2 r$

For ccp, Z = 4 = no. of O-atoms

No. of octahedral voids = 4

No. of tetrahedral voids = 2 $\times$ 4 = 8

No. of Al³⁺ ions = m $\times$ 4

No. of Mg²⁺ ions = n $\times$ 8

Thus, the formula of the mineral is Al_4m Mg_8nO₄

4m(+3) + 8n(+2) + 4($-$2) = 0

12m + 16n $-$ 8 = 0

4(3m + 4n $-$ 2) = 0

3m + 4n = 2

Possible values of m and n are ${1 \over 2}$ and ${1 \over 8}$ respectively.

From the figure, it can be seen that the cation A⁺ occupies octahedral void formed by the anion X^$-$. The radius ratio for an octahedral void is r_A⁺ / r_X^$-$ = 0.414. Now, given that the radius of anion X^$-$ is 250 pm, so the radius of A⁺ is

r_A⁺ = 0.414 $\times$ 250 = 103.50 $ \simeq $ 104 pm

8 X atoms present at the corners.

Atoms contribute to 1 unit cell $ = {1 \over 8} \times 8 = 1$

6 X atoms present at the face centres.

Atoms contribute to 1 unit cell $ = 6 \times {1 \over 2} = 3$

Total X atoms $ = 3 + 1 = 4$

4 M atoms present at edge centres.

Atoms present in 1 unit cell $ = 4 \times {1 \over 4} = 1$

1 M atom present at body centre and it contribute completely to 1 unit cell.

Thus, total M atoms in one unit cell $ = 1 + 1 = 2$

Ratio is $M:X::2:4::1:2$

Thus, empirical formula is $M{X_2}$.

The diagonal is given as 4r, where r is the radius of atom or sphere forming close packed structure.

$ \begin{aligned} \text { Diagonal } & =\sqrt{\mathrm{L}^2+\mathrm{L}^2}=\sqrt{2 \mathrm{~L}^2} \\\\ 4 r & =\sqrt{2 \mathrm{~L}^2} \text { or } \mathrm{L}=\frac{4 r}{\sqrt{2}}=2 \sqrt{2} r \\\\ \text { or } \text { Total area } & =\mathrm{L}^2 \\\\ (2 \sqrt{2} r)^2 & =8 r^2 \end{aligned} $

Number of spheres inside the square is

$ 1+4\left(\frac{1}{4}\right)=2 $

Area of each sphere $=\pi r^2$

Total area of spheres $=2 \times \pi r^2$

$ \begin{aligned} \text { Packing fraction } & =\frac{\text { Total area of spheres }}{\text { Total area }} \\\\ & =\frac{2 \times \pi r^2}{8 r^2}=\frac{\pi}{4}=0.785 \end{aligned} $

So, the percentage fraction is $78.5 \%$.

Frenkel defect occurs in compounds in which the anions are much larger in size than cations. It is a dislocation effect. The presence of Schottky defects lowers the density of the crystal, it hence affects the physical properties of the crystal. In the metal excess defects, the electrons trapped in anion vacancies are known as F-centres.

The number of atoms in a HCP unit cells is 6.

3 atoms are at the middle of the unit cell and each atom contributes 1 atom to the unit cell.

2 atoms are at the two face centres of the unit cell and each atom contributes 1/2 atom to the unit cell.

12 atoms are at the 12 corners of the unit cell and each atom contributes 1/6 atom to the unit cell.

The number of atoms in one unit cell

(3 $\times$ 1) + (2 $\times$ 1/2) + (12 $\times$ 1/6) = 3 + 1 + 2 = 6

Volume of unit cell = Area of base $\times$ Height of the unit cell

The (HCP) is a hexagonal prism structure. The base of (HCP) is a hexagon. We know that the hexagon is made of the 6 equilateral triangles.

Area of equilateral triangle $ = {{\sqrt 3 } \over 4}\,{a^2}$

Therefore, the area of the base is given as,

Area of base $ = 6\times{{\sqrt 3 } \over 4}\,{a^2}$

Let's the 'h' be the height of the hexagonal unit

Therefore, we have

${h \over a} = \sqrt {{8 \over 3}} $

Now, substitute these values in the volumes of (HCP) we have,

Volume of HCP = Area of base $\times$ height of the hexagon

$\therefore$ Volume of HCP $ = 6 \times {{\sqrt 3 } \over 4}\,{a^2} \times \sqrt {{8 \over 3}} a$

Since, spheres touch each other, the edge length 'a' is equal to twice the radius of the atom. That is a = 2r

Substitute these values in the volume we get,

Volume of HCP $ = 6 \times {{\sqrt 3 } \over 4}\,{a^2} \times \sqrt {{8 \over 3}} a$

$ = 6 \times {{\sqrt 3 } \over 4}\,{(2r)^2} \times {{\sqrt 8 } \over {\sqrt 2 }}2r = 24\sqrt 2 {r^3}$

Packing fraction = $\frac{\mathrm{Total~volume~of~unit~cell}}{\mathrm{Volume~occupied}}$

Substituting the values of occupied sphere and volume of the unit cell we have:

$ \Rightarrow {{6 \times {4 \over 3} \times \pi {r^3}} \over {24\sqrt 2 {r^3}}} = 0.74$

Rearranging and solving % occupied space = 74%

The space in the HCP unit cell = 100 $-$ 74 = 26%

(A) Simple cubic and face centered cubic unit cells have cell parameters as $a = b = c$ and $\alpha = \beta = \gamma= 90$

They belong to same crystal system.

(B) Cubic and rhombohedral unit cells have cell parameters $a = b = c$ and $\alpha = \beta = \gamma$. They are two crystal systems.

(C) Cell parameters of cubic unit cell: $a = b = c$ and $\alpha = \beta = \gamma= 90$

b. Rhombohedral unit cell has cell parameters:

$a = b = c$ and $\alpha = \beta = \gamma= 90$ not equal to 90

(D) Hexagonal and monoclinic unit cells are two crystal system:

The have only two crystallography angles of 90$^\circ$.